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3 years ago

6

Driving a piling into a harbor bottom, a pile driver sinks the piling 24 inches on the first stroke, 18 inches on the second str

oke, and 13 1/2 inches on the third stroke. If the sequence is continued, how far will the piling be driven down on the 5th stroke?

1 answer:

ira [324]3 years ago

6 0

Answer:

7.6 inches to the nearest tenth.

Step-by-step explanation:

18/24 = 3/4.

13.5 / 18 = 3/4.

This is a Geometric sequence with common ratio 3/4 or 0.75.

We need to find the fifth term of the sequence.

5th term = a1 (r)^(5 - 1)

= 24 * (0.75)^(5-1)

= 24 * 0.316406

= 7.6 inches.

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Assume that military aircraft use ejection seats designed for men weighing between 141.8 lb and 218 lb. If women's weights are

Mariulka [41]

Answer:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

Step-by-step explanation:

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(173.6,49.8)$

Where $\mu=173.6$ and $\sigma=49.8$

We are interested on this probability

$P(141.8$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(141.8$

And we can find this probability with this difference and using the normal standard table:

$P(-0.639$

Then the answer would be approximately 55.3% of women between the specifications. And that represent more than the half of women

8 0

3 years ago

Refer to the random sample of customer order totals with an average of $78.25 and a population standard deviation of $22.50. a.

zysi [14]

Answer:

a) $78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

b) $78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

c) For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

d) $ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

Step-by-step explanation:

Part a

For this case we have the following data given

$\bar X = 78.25$ represent the sample mean for the customer order totals

$\sigma =22.50$ represent the population deviation

$n= 40$ represent the sample size selected

The confidence level is 90% or 0.90 and the significance level would be $\alpha=0.1$ and $\alpha/2 = 0.05$ and the critical value from the normal standard distirbution would be given by:

$z_{\alpha/2}=1.64$

And the confidence interval is given by:

$\bar X -z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

And replacing we got:

$78.25- 1.64 \frac{22.50}{\sqrt{40}}= 72.416$

$78.25+ 1.64 \frac{22.50}{\sqrt{40}}= 84.084$

Part b

The sample size is now n = 75, but the same confidence so the new interval would be:

$78.25- 1.64 \frac{22.50}{\sqrt{75}}= 73.989$

$78.25+ 1.64 \frac{22.50}{\sqrt{75}}= 82.511$

Part c

For this case when we increase the sample size the margin of error would be lower and then the interval would be narrower

Part d

The margin of error is given by:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

Solving for n we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

And replacing the info we have:

$n=(\frac{1.640(22.50)}{5})^2 =54.46 \approx 55$

3 0

3 years ago

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bixtya [17]

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Question
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17x³ + (3x + 8x³<span>)

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Take out common factor
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x(25x² + 3)

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Answer: x(25x² + 3)
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